If $\mathfrak a\subset A$ is a finitely generated ideal, and if $\mathfrak p$ is a prime ideal, then $S(\mathfrak a)\cap\mathfrak p\not=\emptyset ?$ For an ideal $\mathfrak a\subset A,$ define $S(\mathfrak a)=\{f\in A\mid f\not\in x, \forall x\in D(\mathfrak a)\};$ namely, $S(\mathfrak a)$ is the set of elements that do not belong to any prime ideal which does not contain $\mathfrak a.$
Then the question is:  

For which ring $A$ is the following true?
  If $\mathfrak a$ is finitely generated and if $\mathfrak p$ is a prime ideal containing $\mathfrak a,$ then  $S(\mathfrak a)\cap\mathfrak p\not=\emptyset.$  

My attempt:
When $\mathfrak a$ is generated by a single element $f,$ then clearly $f\in S(\mathfrak a)\cap\mathfrak p.$
By induction on the number of generators of $\mathfrak a=\left(f_1,\cdots,f_n\right),$ I found elements  $x_1,\cdots,x_n$ such that $x_i\in S(\mathfrak a_i)\cap\mathfrak p, \forall i,$ where $\mathfrak a_i=\left(f_1,\cdots,\hat f_i,\cdots,f_n\right).$ And, if some $x_i\in S(f_i),$ then this $x_i$ belongs to the intersection in question, so we assume that $x_i\not\in S(f_i).$
Then we have $\alpha_i:=x_1\cdots\hat x_i\cdots x_n\not\in S(f_j), \forall j\not=i,$ and $\alpha_i\in S(f_i).$ Thus I tried to somehow glue together these $\alpha_i$ to produce one element in the intersection in question, to no avail.
However, if there are only finitely prime ideals of $A,$ then I managed to produce a weighted sum of these $\alpha_i$ which does belong to $S(\mathfrak a)\cap\mathfrak p.$  
P.S. My attemps above ignore completely the prime-ness of $\mathfrak p,$ and is valid for every ideal containing $a,$ which led me to believe that I must have missed something.  
Any help will be greatly appreciated; thanks in advance.  
Edit:
Thanks to Jake Levinson, I now know that the proposition is not true when $A$ is Noetherian of dimension $\ge2,$ and I have shown its validity when $A$ is a PID or when $A$ has finitely many prime ideals. So the remaining case is when $\operatorname{dim}(A)=1,$ and $A$ is not a PID, and has infinitely many prime ideals.
 A: This is not true. For example, let $A$ be Noetherian and let $\mathfrak a$ be an ideal of height $2$ or more. So in this case the set $D(\mathfrak a)$ of "prime ideals which do not contain $\mathfrak a$" includes every prime ideal of height $1$ (or $0$).
In particular if $f \in S(\mathfrak a)$, then $f$ is a unit. So $S(\mathfrak a) \cap \mathfrak{p} = \emptyset$ for any proper ideal $\mathfrak{p}$.
As an example, let $\mathfrak a = (x,y)$ in $k[x,y]$.
Note: It's worth pointing out that the cases you found proofs for are somewhat exceptional, -- principal ideals, and rings with only finitely-many prime ideals. If $A$ is assumed Noetherian, then principal ideals have height at most $1$, and the "finitely-many prime ideals" condition implies $\dim A \leq 1$.
As per your request, I will consider the case where $\mathfrak{a}$ is itself a prime ideal, of height one (and $A$ is Noetherian). In this case it follows that $S(\mathfrak{a}) \cap \mathfrak{p} = S(\mathfrak{a}) \cap \mathfrak{a}$, for any choice of $\mathfrak{p} \supseteq \mathfrak{a}$.  (Note that every non-unit is contained in some height-one prime ideal). In particular $f \in S(\mathfrak{a}) \cap \mathfrak{a}$ if and only if $\mathfrak{a}$ is the unique height-one prime ideal containing $f$, that is, $\mathrm{div}(f) = n \cdot \mathfrak{a}$ for some $n\geq 1$. Equivalently, $\mathfrak{a}$ is a torsion element in the class group of $A$. (Such elements are rare in general.)
Sometimes these elements $f$ will exist for all ideals $\mathfrak{a}$ of $A$: for example, if $A$ is the ring of integers of a number field, then its class group is finite, so some power of $\mathfrak{a}$ is a principal ideal.
On the other hand, there are counterexamples even if $A$ is still taken to be a Dedekind domain -- just take one whose class group has elements of infinite order, For instance, let $A = \mathbb{C}[x,y]/(y^2 - x^3 - 1)$. There are ideals $\mathfrak{a}$ in $A$ no power of which is principal (in fact, such ideals are dense in $\mathrm{Spec}(A)$.) But $f \in S(\mathfrak{a}) \cap \mathfrak{a}$ implies that $\sqrt{(f)} = \mathfrak{a}$, hence by ideal factorization $(f) = \mathfrak{a}^n$ for some $n$.
